show that the curve:

γ(t)=[(1+tē)/t),t+1,(1-t)/t)]

is planar

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- Aug 4th 2010, 01:59 AMulysses123planar curve
show that the curve:

γ(t)=[(1+tē)/t),t+1,(1-t)/t)]

is planar - Aug 4th 2010, 02:52 AMAckbeet
What ideas have you had so far?

- Aug 4th 2010, 03:29 AMulysses123
i am not sure what this question is asking so im am not sure how to procced.To start with What is a planar curve?

- Aug 4th 2010, 04:40 AMAckbeet
If you think of all of three-dimensional space, and curves that can trace their paths out in that space, a planar curve is one of those paths such that the entire path lies in one plane. It never deviates from the plane. Does that make sense?

How could you characterize a plane? - Aug 4th 2010, 04:52 AMulysses123
if i solve this eqn and turn it into cartesian coordinates i get the eqn of a plane:z=x-y,

but i still don't understand the question.Is it saying whether or not there existst some plane that contains all the points?

because it is actually an eqn of a plane. - Aug 4th 2010, 05:02 AMAckbeet
No, the equation

$\displaystyle \displaystyle{\vec{Y}(t)=\left\langle \frac{1+t^{2}}{t},t+1,\frac{1-t}{t}\right\rangle}$

cannot describe a plane, because a plane has two degrees of freedom, and this equation only has one degree of freedom $\displaystyle (t)$. It's a space curve.

Quote:

Is it saying whether or not there exists some plane that contains all the points?

1. Write down an equation that characterizes a plane, any plane, in 3D space.

2. Figure out how to characterize vectors that "lie alongside" the curve.

Once you've done both of these steps, your task is to get your representation in #2 to look like #1. Then you're done.

So, how about #1 there? How can you characterize all the points in a plane? - Aug 4th 2010, 05:25 AMulysses123
ax+by+cz=d

I guess if the curves are parrallel they must be some scalar multiple but im not really seing how this will help me find the cartesian eqn to see that it lies in a two dimensional plane - Aug 4th 2010, 05:36 AMAckbeet
Another form of your plane equation is this (it's entirely equivalent to the one you gave):

$\displaystyle \vec{n}\cdot(\vec{x}-\vec{x}_{0})=0,$

where $\displaystyle \vec{x}$ varies, and $\displaystyle \vec{x}_{0}$ is constant. The vector $\displaystyle \vec{n}$ is normal to the surface of the plane, and all the vectors $\displaystyle \vec{x}-\vec{x}_{0}$ are*in the plane*. This is the geometric interpretation that I think is significant for your problem. Incidentally, if you take my equation and do this:

$\displaystyle \vec{n}\cdot\vec{x}-\vec{n}\cdot\vec{x}_{0}=0,$ then you can chuck the constant term over to the other side:

$\displaystyle \vec{n}\cdot\vec{x}=\vec{n}\cdot\vec{x}_{0},$

and get your equation with $\displaystyle d=\vec{n}\cdot\vec{x}_{0}.$

I think my original equation there will be the most useful for you. So that's my #1 step. How do you propose to do #2? I should point out that there are at least two ways to do #2. If you know calculus, there's a quicker way. Otherwise, you can still do the (somewhat more lengthy) algebra approach. - Aug 4th 2010, 05:58 AMulysses123
if i multiply each of the parametric equations by a constant and use the form for a plane i get n=a+c,a+b-c

and d=-b and the vector x gives me the components 1 /t,t - Aug 4th 2010, 06:46 AMAckbeet
I'm not following your logic here. What are you trying to do? Why would you multiply each of the parametric equations by a constant? Vectors from the origin to points in a plane are not themselves in the plane, unless the plane contains the origin. That is definitely not the general case. You can see at a glance that the origin is not contained in the plane we're dealing with, because the first component is never zero.

You need to have a way of generating a representation of all the vectors*in the plane*. Think about this geometrically. You have a space curve. You want to show that all the points in the curve are in a plane. So, somehow, you have to generate a whole bunch of vectors in the suspected plane. But a vector is not the same thing as a point. How can you get vectors out of points? - Aug 6th 2010, 05:16 AMulysses123
i've given up on this question and moved on because i cant seem to figure it out!

- Aug 6th 2010, 05:29 AMAckbeetQuote:

...i've given up on this question and moved on...

If the first is the situation, then I would urge you to re-read my earlier posts. This is a do-able problem, and my hints should provide at least some direction. Your posts have indicated that you're really not paying attention to my hints. That would be perfectly fine, if you were on to something yourself. But your posts are all confused. So, if you really do want to solve this problem, take another look at my hints, and pepper me with specific questions about what you don't understand.

If the second is the situation, then I'm afraid I can't help you any more. This forum is for people who want just that little bit of help they need to get unstuck. It's not for people who simply want answers to their problems. We want you to understand mathematics. Remember proverb by Confucius:

Tell me and I forget. Show me and I remember. Involve me and I understand.

Regards. - Aug 6th 2010, 06:29 PMulysses123
i have tried it. i''ll try it again from scratch and post a question if i still cant figure it out.Because i really do want to solve this, but its doing my head in.

- Aug 6th 2010, 07:23 PMulysses123
what i have done is take three arbitrary t values,t=1,2,3.Then use these point to solve the eqn of a plane.This gives me n=(a,-a,-a) and d=0.

This satisfies all points that are part of the parametric curve.Does this mean that ALL the points of the parametric curve lie in this plane, but this eqn of a plane is not a representation of the parametric curve itself, only the points that lie on the parametric curve? - Aug 7th 2010, 07:01 AMAckbeet
Unfortunately, your "arbitrary" t values aren't arbitrary enough. Here are two approaches, and they both work:

1. Show that all vectors of the form $\displaystyle Y(t_{1})-Y(t_{2})$ are perpendicular to some constant vector.

2. Show that all vectors of the form $\displaystyle Y'(t)$ are perpendicular to some constant vector.

Can you see why these approaches will get you what you need?